Question

Use the Divergence Theorem to calculate the flux of
F(x,y,z)=(x^5 +y^5 +z^5 -2x-3y-4z)i+sin(2y)j+4z(sin(y))^2 k
across the surface of the tetrahedron bounded by the coordinate planes and the plane
x+y+z=1.

Answer 1

\[\iint\limits_{S}\vec{F}\bullet d\vec{S}~=~\iiint \limits_{E}\text{div}(\vec{F}) dV \]

Answer 2

work the divergence and plug it in

Answer 3

we can write this as a volume integral over \(E=\{(x,y,z)\in\mathbb{R}^3:x+y+z\le 1, x\ge 0, y\ge0, z\ge0\}\) using the divergence theorem as ganeshie8 pointed out, which is something like: $$\iint\limits_S F\cdot dS=\iiint\limits_E\nabla\cdot F\ dV$$

Answer 4

\[\int\limits_{0}^{1}\int\limits_{0}^{1-x}\int\limits_{1}^{1-x-y}dzdydx\]

Answer 5

now, using Cartesian coordinates so \(dV=dz\ dy\ dx\) $$\int_0^1\int_0^{1-x}\int_0^{1-x-y}\nabla\cdot F\ dz\ dy\ dx$$
clearly \(\nabla\cdot F=(5x^4-2) +2\cos(2y)+4\sin^2(y)\) so: $$\int_0^1\int_0^{1-x}\int_0^{1-x-y}\left(5x^4-2+2\cos(2y)+4\sin^2(y)\right)\,dz\,dy\,dx$$

Answer 6

can you tell me if I have the correct limits

Answer 7

no, \(0\le z\le 1-x-y\)

Answer 8

Thanks for the help, I had a typo with the dz limit

Answer 9

At this point, you may use wolfram to evaluate the triple integral for you

http://www.wolframalpha.com/input/?i=%5Cint_0%5E1%5Cint_0%5E%7B1-x%7D%5Cint_0%5E%7B1-x-y%7D%5Cleft%285x%5E4-2%2B2%5Ccos%282y%29%2B4%5Csin%5E2%28y%29%5Cright%29%5C%2Cdz%5C%2Cdy%5C%2Cdx